Two positive point charges , each `Q` , are fixed at separation `d`. A third charge `q` is placed in the middle. Describe the equilibrium of the third charge.

Explanation

If `q` is positive
(a) Net force on `q` is zero , hence it is in equilibrium . Let `q` be displaced slightly towards right by `x`.
Force on `q`:
Due to `Q at A , F_(1) = (1)/( 4pi in_(0)) (Qq)/(((d)/(2) - x)^(2))` , towards left
Due to `Q at B , F_(1) = (1)/( 4pi in_(0)) (Qq)/(((d)/(2) + x)^(2))` , towards right
`F_(1) gt F_(2)` , net force on `q` is towards left. Hence it will bring `q` to equilibrium position , and equilibrium is stable along the line `AB`.
(b) Let `q` be placed slightly perpendicular to line `AB` , as shown. The resultant force on `q` will be away from `O` and the particle will not return to its equilibrium position `O`, hence unstable equilibrium.
If `q` is negative
(a) If `-q` is displaced slightly towards right , the net force on it will be away from origin i.e. along `OA` and the particle will not return to its equilibrium position , hence unstable equilibrium.
(b) If `-q` is placed slightly perpendicular to line AB, the resultant force on `-q` will be towards equilibrium position , hence this force will bring `-q` to `O`, hence stable equilibrium.